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Nataliya [291]
3 years ago
7

At a small university, there are 1800 female students and 2000 male students. Which of the following expresses the ratio of fema

le students to male students?
A.) 10 to 9
B.) 9/10
C.) 10/9
D.) 10 : 9
Mathematics
2 answers:
Bas_tet [7]3 years ago
6 0
To solve you would turn this into a fraction and simplify. An easier way on this problem is to look at the answers, they are all the same exact thing except for B. I am pretty certain B. is the correct answer.
weqwewe [10]3 years ago
6 0

Answer:  Option 'B' is correct.

Step-by-step explanation:

Since we have given that

Number of female students = 1800

Number of male students = 2000

Ratio of female students to male students is given by

Female students : Male students

1800:2000\\\\=9:10\\\\=\dfrac{9}{10}

Hence, Option 'B' is correct.

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HELP PLZZZZZZZ STANDARD FORM
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Answer:

-12n^{2} + 12

Step-by-step explanation:

h(n) = -4n + 4           (h o g) (n)

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7 0
3 years ago
Ricky is packing food for camping trip. He has 24 granola bars and 30 protein bars. Ricky wants to make identical bags. Each wil
dimaraw [331]

Answer:

6

Step-by-step explanation:

first, we need to find the GCF (Greatest Common Factor) which is the highest number that can go into every number in the equation (In this case, those numbers are 24 and 30)

The GCF for this would be 6, meaning that there would be 6 bags with and equal amount of granola bars and protein bars.
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3 0
2 years ago
Find dy/dx by implicit differentiation for ysin(y) = xcos(x)
tatyana61 [14]

Answer:

\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}

Step-by-step explanation:

So we have:

y\sin(y)=x\cos(x)

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]

Let's do each side individually.

Left Side:

We have:

\frac{d}{dx}[y\sin(y)]

We can use the product rule:

(uv)'=u'v+uv'

So, our derivative is:

=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]

We must implicitly differentiate for y. This gives us:

=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]

For the sin(y), we need to use the chain rule:

u(v(x))'=u'(v(x))\cdot v'(x)

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})

Simplify:

=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}

And we are done for the right.

Right Side:

We have:

\frac{d}{dx}[x\cos(x)]

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]

Differentiate:

=\cos(x)-x\sin(x)

So, our entire equation is:

=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}

And we're done!

5 0
3 years ago
Solve 12 2/3 divide 1/3
Alina [70]
Your answer is 38 I hope its correct :)
7 0
3 years ago
Read 2 more answers
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